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Thepotemich [5.8K]
3 years ago
6

A flowerpot falls off a balcony 85m above the street how long does it take to hit the ground

Physics
2 answers:
GenaCL600 [577]3 years ago
7 0
It will take about 4.2 seconds
Ksivusya [100]3 years ago
7 0
The distance a falling object falls in some amount of time is

        D = 1/2  a  T²

If this flowerpot falls off a balcony on Earth, then 'a' is the
acceleration of gravity on Earth, and we can write

      85 m  =  1/2 (9.8 m/s²) T²

Divide each side by  4.9 m/s² :

      85/4.9  s²  =  T²

Square root each side:

      T  =  √(85/4.9)  seconds

          =      4.165 seconds .
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Solution :

Energy of photon, E = 6.7 eV

                              E = $6.7 \times 1.602 \times 10^{-7}$ joule

Kinetic energy, $K.E. =\frac{1}{2} mv^2 = 1.602 \times 6.7 \times 10^{-7}$

$v^2=\frac{2 \times 1.602 \times 6.7 \times 10^{-7}}{1.6726 \times 10^{-27}}$

   $=12.834 \times 10^{-20}$

Kinetic energy at high speeds

$(r-1)\times mc^2 = 6.7 \ eV$

$(r-1)=\frac{6.7 \times 1.602 \times 10^{-7}}{1.6726 \times 10^{-27} \times 9 \times 10^{16}}$

r - 1 = 7130

r = 7130 + 1

r  = 7131

$\frac{1}{\sqrt{1-\frac{v^2}{C^2}}}=7131$

$1-\frac{v^2}{C^2} = \left(\frac{1}{7131}\right)^2$

$v^2=C^2\left[1-\left(\frac{1}{7131}\right)^2\right]$

$v=0.99999999017C$

Δ = 1 - 0.99999999017

   = 0.00000000933

Relative mass, $m_{rel}=r.m$

                                $=7131 \times 1.6728 \times 10^{-27}$

                               $=1.1927 \times 10^{-23}$ kg

                                 

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A battery charges a parallel-plate capacitor fully and then is removed. The plates are immediately pulled apart. (With the batte
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Answer:

<em>There will be an increase in potential difference.</em>

Explanation:

As we know that the potential difference depends upon the capacitance.

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When battery is disconnected the charge remains constant on the plates but the capacitance decreases. As the capacitance has an inverse relation with the potential difference, there will be an increase in it.

In addition to that the potential difference can also be defined as the product of field and distance between the plates. As the charge is constant so the field is constant. Upon increasing the separation between the plates the potential difference will also increased.

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